The geometric mean of two numbers is 6 and th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the two numbers be $a$ and $b$.Given that the geometric mean is $6$,we have $\sqrt{ab} = 6$,which implies $ab = 36$.Given that the arithmetic

Vedclass · Accepted Answer

$(4, 9)$

Vedclass · Answer

(B) Let the two numbers be $a$ and $b$.Given that the geometric mean is $6$,we have $\sqrt{ab} = 6$,which implies $ab = 36$.Given that the arithmetic mean is $6.5$,we have $\frac{a+b}{2} = 6.5$,which implies $a+b = 13$.We can form a quadratic equation $x^2 - (a+b)x + ab = 0$ to find the numbers.Substituting the values,we get $x^2 - 13x + 36 = 0$.Factoring the quadratic equation: $x^2 - 9x - 4x + 36 = 0$.$x(x - 9) - 4(x - 9) = 0$.$(x - 4)(x - 9) = 0$.Thus,the numbers are $4$ and $9$.

The geometric mean of two numbers is $6$ and their arithmetic mean is $6.5$. The numbers are

Similar Questions

If the arithmetic mean of two numbers $a$ and $b$,where $a > b > 0$,is five times their geometric mean,then $\frac{a + b}{a - b}$ is equal to

If $A$ and $G$ are the $A.M.$ and $G.M.$ respectively between two positive numbers,prove that the numbers are $A \pm \sqrt{(A + G)(A - G)}$.

If the ratio of $H.M.$ and $G.M.$ between two numbers $a$ and $b$ is $4:5$,then the ratio of the two numbers will be

If $A$ and $G$ are the arithmetic mean and geometric mean of $a$ and $b$ respectively,then what is the value of $A - G$?

Consider two positive numbers $a$ and $b$. If the arithmetic mean of $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and the geometric mean of $a$ and $b$ exceeds their harmonic mean by $\frac{6}{5}$,then the absolute value of $(a^2 - b^2)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module